The Student Room Group

Special Relativity - Energy

I just want to clear something up that has been bugging me lately. So E=mc2=mo(gamma)c2.

Say a proton is accelerated from rest to some relativistic speed. It's gain in energy is only from a gain in kinetic energy which can be found by the mass-energy equivalence principal. So is it correct to say that ET=mo(gamma)c2 such that Ek= mo(gamma)c2 - moc2

mo=Rest mass
ET=Total energy
EK=Kinetic energy of proton

Scroll to see replies

Original post by Protoxylic
I just want to clear something up that has been bugging me lately. So E=mc2=mo(gamma)c2.

Say a proton is accelerated from rest to some relativistic speed. It's gain in energy is only from a gain in kinetic energy which can be found by the mass-energy equivalence principal. So is it correct to say that ET=mo(gamma)c2 such that Ek= mo(gamma)c2 - moc2

mo=Rest mass
ET=Total energy
EK=Kinetic energy of proton


If ET = mo(gamma)c2 and Ek= mo(gamma)c2 - moc2, so it must be Ek= ET- moc2.

So if I understand you well do you think that the kinetic energy of a proton is the difference between the total energy of this proton and the rest mass in relation to the light speed?
(edited 9 years ago)
Reply 2
Original post by Kallisto
If ET = mo(gamma)c2 and Ek= mo(gamma)c2 - moc2, so it must be Ek= ET- moc2.

So if I understand you well do you think that the kinetic energy of a proton is the difference between the total energy of this proton and the rest mass in relation to the light speed?


Basically, the increase in the energy of the proton is the kinetic energy. And this increase is the difference between the energy equivalence of mass at speed and at rest.
Original post by Protoxylic
Basically, the increase in the energy of the proton is the kinetic energy. And this increase is the difference between the energy equivalence of mass at speed and at rest.


I have thought about your equation.

Every single particle (proton in this case) has a total energy. And this Energy is turned into kinetic Energy whenever the proton is moving. The rest mass is the one when the mass of a particle is at rest, but this mass can be increased by increasing velocity. Thus this rest mass is at the maximum when it gets the light speed, as nothing is faster than light.

We know that whenever energy is transformed, a sum of energy of a kind of energy (total energy in this case) gets lost by transferring this sum of energy to an other kind of energy (kinetic energy in this case). And this loss of total energy is caused by the increasing mass and increasing velocity when the proton is moving what leads to kinetic energy.

And that is to say, the kinetic energy is increased when the total energy is decreased.

So from this point of view it seems to be plausible that the kinetic energy is the difference between the total energy and the increasing rest mass to increasing velocity (up to light speed).
(edited 9 years ago)
Reply 4
Original post by Kallisto
I have thought about your equation.

Every single particle (proton in this case) has a total energy. And this Energy is turned into kinetic Energy whenever the proton is moving. The rest mass is the one when the mass of a particle is at rest, but this mass can be increased by increasing velocity. Thus this rest mass is at the maximum when it gets the light speed, as nothing is faster than light.

We know that whenever energy is transformed, a sum of energy of a certain kind of energy (total energy in this case) gets lost by transferring it to an other kind of energy (kinetic energy in this case). And this loss of total energy is caused by the increasing mass and increasing velocity when the proton is moving what leads to kinetic energy.

And that is to say, the kinetic energy is increased when the total energy is decreased.

So from this point of view it seems to be plausible that the kinetic energy is the difference between the total energy and the increasing rest mass to increasing velocity (up to light speed).


Are you sure about that statement? It seems weird to say that EK increases when ET decreases. So ET=moc2+KE which is to say that the total energy is the rest energy + kinetic energy of the particle. So since ET in this case is mo(gamma)c2 because if work is done on the particle to accelerate it to a speed v, then this energy is stored within the particle as kinetic energy. And this energy has a mass equivalence to it such that the particle gains mass equal to EK/c2
Original post by Protoxylic
I just want to clear something up that has been bugging me lately. So E=mc2=mo(gamma)c2.

Say a proton is accelerated from rest to some relativistic speed. It's gain in energy is only from a gain in kinetic energy which can be found by the mass-energy equivalence principal. So is it correct to say that ET=mo(gamma)c2 such that Ek= mo(gamma)c2 - moc2

mo=Rest mass
ET=Total energy
EK=Kinetic energy of proton


For starters, the concept of 'kinetic energy' in special relativity is with reference to a specific frame.

mo is something that should just be scrapped... take m as the rest mass.

There is a frame invariant quantity which is m2c4 = E2 - p2c2 which is Einstein's famous mass-energy relation and for a particle at rest reduces to E=mc2
Reply 6
Original post by natninja
For starters, the concept of 'kinetic energy' in special relativity is with reference to a specific frame.

mo is something that should just be scrapped... take m as the rest mass.

There is a frame invariant quantity which is m2c4 = E2 - p2c2 which is Einstein's famous mass-energy relation and for a particle at rest reduces to E=mc2


So I have just done a past paper question regarding a very similar version of what I stated. They used EK=mo(gamma)c2-moc2. Meaning that the kinetic energy is the difference between the mass-energy equivalence at speed and the rest mass, which makes sense because this would give you the work done on the particle, since it is it's change in energy, and this work would just be converted to kinetic energy.
Original post by Protoxylic
So I have just done a past paper question regarding a very similar version of what I stated. They used EK=mo(gamma)c2-moc2. Meaning that the kinetic energy is the difference between the mass-energy equivalence at speed and the rest mass, which makes sense because this would give you the work done on the particle, since it is it's change in energy, and this work would just be converted to kinetic energy.


Kinetic energy absolutely is the difference between total energy and rest energy. There are a few ways of expressing this relation, but that is the simplest, if not necessarily the most useful.

Ek=(γ1)moc2E_k = (\gamma - 1) m_o c^2
Original post by Protoxylic
Are you sure about that statement? It seems weird to say that EK increases when ET decreases. So ET=moc2+KE which is to say that the total energy is the rest energy + kinetic energy of the particle. So since ET in this case is mo(gamma)c2 because if work is done on the particle to accelerate it to a speed v, then this energy is stored within the particle as kinetic energy. And this energy has a mass equivalence to it such that the particle gains mass equal to EK/c2


Now that you said it in a way like that, it seems to be silly what I stated. I guess it was very naive to give an explanation to an equation of quantum physics by considering about the law of conservation energy.

If the total Energy is stored in a particle as a kinetic Energy, so my whole consideration was wrong.
(edited 9 years ago)
Original post by Protoxylic
So I have just done a past paper question regarding a very similar version of what I stated. They used EK=mo(gamma)c2-moc2. Meaning that the kinetic energy is the difference between the mass-energy equivalence at speed and the rest mass, which makes sense because this would give you the work done on the particle, since it is it's change in energy, and this work would just be converted to kinetic energy.


Yes, though it's frame dependent, Kinetic energy isn't really a thing... it is always perfectly possible to find a frame in which the kinetic energy is zero.

You know how you found out that some things were simplifications at GCSE level? The same is true at A-level and this is one of them.
Original post by natninja
Yes, though it's frame dependent, Kinetic energy isn't really a thing... it is always perfectly possible to find a frame in which the kinetic energy is zero.

You know how you found out that some things were simplifications at GCSE level? The same is true at A-level and this is one of them.


So in a hypothetical situation where the proton was accelerated in a straight line from a particle accelerator. The kinetic energy relative to the stationary particle accelerator is given by the difference in it's total energy and it's rest energy.
Original post by Kallisto
I have thought about your equation.

Every single particle (proton in this case) has a total energy. And this Energy is turned into kinetic Energy whenever the proton is moving. The rest mass is the one when the mass of a particle is at rest, but this mass can be increased by increasing velocity. Thus this rest mass is at the maximum when it gets the light speed, as nothing is faster than light.

We know that whenever energy is transformed, a sum of energy of a kind of energy (total energy in this case) gets lost by transferring this sum of energy to an other kind of energy (kinetic energy in this case). And this loss of total energy is caused by the increasing mass and increasing velocity when the proton is moving what leads to kinetic energy.

And that is to say, the kinetic energy is increased when the total energy is decreased.

So from this point of view it seems to be plausible that the kinetic energy is the difference between the total energy and the increasing rest mass to increasing velocity (up to light speed).


Light speed can never be attained by a particle with mass. As its speed tends to the speed of light, its mass* would tend to infinity. There is no maximum.

*By which we really mean energy. Rest mass obviously never changes.
(edited 9 years ago)
Original post by Kallisto
Now that you said it in a way like that, it seems to be silly what I stated. I guess it was very naive to give an explanation to an equation of quantum physics by considering about the law of conservation energy.

If the total Energy is stored in a particle as a kinetic Energy, so my whole consideration was wrong.


1) It isn't quantum physics

2) Actually there was only one thing wrong with your consideration which was that you said that the rest mass increases. The rest mass is a constant and is frame independent. The difference between the rest energy and the total energy is the 'kinetic' energy the particle has IN THAT FRAME.
Original post by Protoxylic
So in a hypothetical situation where the proton was accelerated in a straight line from a particle accelerator. The kinetic energy relative to the stationary particle accelerator is given by the difference in it's total energy and it's rest energy.


That is true. However, when it comes to quantifying kinetic energy you need to specify what frame the observer is in. Otherwise it is equally valid to say that the proton has zero kinetic energy and it is the accelerator that is moving.

Fun problem for you to think about - can pair production happen in a vacuum?
Original post by natninja
That is true. However, when it comes to quantifying kinetic energy you need to specify what frame the observer is in. Otherwise it is equally valid to say that the proton has zero kinetic energy and it is the accelerator that is moving.

Fun problem for you to think about - can pair production happen in a vacuum?


Well I assume that this situation is just a photon in a vacuum and in this case you would need a nucleus to interact with, so no in that respect. But taking a vacuum to be a true vacuum except the nucleus the photon is interacting with and taking conservation of momentum+energy. Momentum should be conserved so if the photon interacts with the nucleus, the total momentum before must be equal to the total momentum after. You told me that you could take any frame of reference such that the momentum is different in that frame of reference. If you take one of the created particles to be the frame of reference, the the total momentum after wouldn't be zero, since one particle has momentum in one direction and the other has zero momentum. So it probably can't happen.
Original post by Protoxylic
Well I assume that this situation is just a photon in a vacuum and in this case you would need a nucleus to interact with, so no in that respect. But taking a vacuum to be a true vacuum except the nucleus the photon is interacting with and taking conservation of momentum+energy. Momentum should be conserved so if the photon interacts with the nucleus, the total momentum before must be equal to the total momentum after. You told me that you could take any frame of reference such that the momentum is different in that frame of reference. If you take one of the created particles to be the frame of reference, the the total momentum after wouldn't be zero, since one particle has momentum in one direction and the other has zero momentum. So it probably can't happen.


Nah it can (and does) happen in the presence of other matter to interact with (a small virtual photon links the systems)

Try to think of why it would require a nucleus to interact with for a photon in a vacuum.
Original post by natninja
Nah it can (and does) happen in the presence of other matter to interact with (a small virtual photon links the systems)

Try to think of why it would require a nucleus to interact with for a photon in a vacuum.


Well it would be perfectly acceptable to happen because of the other piece of matter acting as a frame of reference such that momentum is conserved?
Original post by Protoxylic
Well it would be perfectly acceptable to happen because of the other piece of matter acting as a frame of reference such that momentum is conserved?


Classical momentum conservation is slightly warped in SR and is really only relevant in one frame (hint) what is conserved between frames are things called Lorentz scalars.
Original post by natninja
Classical momentum conservation is slightly warped in SR and is really only relevant in one frame (hint) what is conserved between frames are things called Lorentz scalars.


So it would be acceptable to say that momentum is conserved when you're observing the pair production from the reference frame of the matter that the photon interacted with? I'm confused as to the hint is preceding the word or after
Original post by Protoxylic
So it would be acceptable to say that momentum is conserved when you're observing the pair production from the reference frame of the matter that the photon interacted with? I'm confused as to the hint is preceding the word or after


Not even. It would be acceptable to say that the total 4-momentum in the centre of momentum/mass frame of the system is zero.

The reason the nucleus is required is that the velocity of the centre of momentum frame doesn't change. But the problem here is where is the centre of momentum of a photon? literally the other matter being in the system gives the system a frame in which it can exist.

Quick Reply

Latest